MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 added 254 characters in body

General pushouts, no; pushouts of pairs of epis, yes. In fact, yes to pushouts of pairs where only one of the arrows of the pair is epi.

It's very easy to see the answer is no for general pushouts: since a coproduct $A + B$ is the pushout of a pair $A \leftarrow 0 \to B$, and since the squaring functor preserves $0$, the squaring functor would preserve this pushout only if it preserved the coproduct. But we all know that for finite cardinalities, $a^2 + b^2$ is generally not equal to $(a+b)^2$; therefore squaring cannot preserve coproducts.

But in general, we can say that for the category of sets, given a pair of functions

$$A \stackrel{f_1}{\leftarrow} P \stackrel{f_2}{\to} B,$$

the canonical arrow

$$\phi: A^2 +_{P^2} B^2 \to (A +_P B)^2$$

is monic. (Lemma to be proved.) Assuming this, suppose for example that $f_2: P \to B$ is epic. Then the pushout of $f_2$ along $f_1$ is also epic (well-known fact); let $A \to C$ denote this epi (so $C$ is shorthand for $A +_P B$). Then $A^2 \to C^2$ is also epic, and since this obviously factors as

$$A^2 \to A^2 +_{P^2} B^2 \stackrel{\phi}{\to} C^2$$

we see $\phi$ is also epi. But it is monic by the lemma, hence $\phi$ is an isomorphism, as we wanted to show.

To prove the lemma, it helps to have a clear picture of how pushouts are formed in $Set$. The pushout $C$ is the set of equivalence classes on $A + B$ where $x \in A + B$ is deemed equivalent to $x' \in A + B$ iff there is a zig-zag path

$$x = x_0 \stackrel{f_{i_1}}{\leftarrow} p_0 \stackrel{f_{i_2}}{\to} x_1 \leftarrow \ldots x_{n-1} \stackrel{f_{i_{n-1}}}{\leftarrow} p_{n-1} \stackrel{f_{i_n}}{\to} x_n = x'$$

where for each $k$, either $p_k$ belongs to $P$ and the arrows out of $p_k$ alternate between $f_1$ and $f_2$, or we are in a "holding pattern" where $p_k$ belongs to $A$ or $B$ and the two arrows out of $p_k$ are both identities. Now it is not hard to convince yourself that given $(x, y)$ in $A^2$ or $B^2$, and $(x', y')$ in $A^2$ or $B^2$, if there is a zig-zag path from $x$ to $x'$, and a zig-zag path from $y$ to $y'$, then there is a zig-zag path from $(x, y)$ to $(x', y')$ with respect to the pair of maps

$$A \times A \stackrel{f_1 \times f_1}{\leftarrow} P \times P \stackrel{f_2 \times f_2}{\to} B \times B;$$

all we do is pair together zig-zag paths in the separate $x$- and $y$-components. (Note that if the zig-zag to get from $y$ to $y'$ is longer than the zig-zag from $x$ to $x'$, we can always insert a holding pattern in the $x$-component so that the zig-zag in the $y$-component can "catch up", i.e., so that the lengths of the zig-zags match up.) But this means precisely that the map $\phi$ is monic.

Edit: In answer to vincenzoml's questions in the comments, I wrote up a proof of the more general desired result which applies to a general $\infty$-pretopos, here.

2 added 494 characters in body

General pushouts, no; pushouts of pairs of epis, yes. In fact, yes to pushouts of pairs where only one of the arrows of the pair is epi.

It's very easy to see the answer is no for general pushouts: since a coproduct $A + B$ is the pushout of a pair $A \leftarrow 0 \to B$, and since the squaring functor preserves $0$, the squaring functor would preserve this pushout only if it preserved the coproduct. But we all know that for finite cardinalities, $a^2 + b^2$ is generally not equal to $(a+b)^2$; therefore squaring cannot preserve coproducts.

But in general, we can say that for the category of sets, given a pair of functions

$$A \stackrel{f_1}{\leftarrow} P \stackrel{f_2}{\to} B,$$

the canonical arrow

$$\phi: A^2 +_{P^2} B^2 \to (A +_P B)^2$$

is monic. (Lemma to be proved.) Assuming this, suppose for example that $f_2: P \to B$ is epic. Then the pushout of $f_2$ along $f_1$ is also epic (well-known fact); let $A \to C$ denote this epi (so $C$ is shorthand for $A +_P B$). Then $A^2 \to C^2$ is also epic, and since this obviously factors as

$$A^2 \to A^2 +_{P^2} B^2 \stackrel{\phi}{\to} C^2$$

we see $\phi$ is also epi. But it is monic by the lemma, hence $\phi$ is an isomorphism, as we wanted to show.

To prove the lemma, it helps to have a clear picture of how pushouts are formed in $Set$. The pushout $C$ is the set of equivalence classes on $A + B$ where $x \in A + B$ is deemed equivalent to $x' \in A + B$ iff there is a zig-zag path

$$x = x_0 \stackrel{f_{i_1}}{\leftarrow} p_0 \stackrel{f_{i_2}}{\to} x_1 \leftarrow \ldots x_{n-1} \stackrel{f_{i_{n-1}}}{\leftarrow} p_{n-1} \stackrel{f_{i_n}}{\to} x_n = x'$$

where for each $k$, either $p_k$ belongs to $P$ and the arrows out of $f_{i_k}$ p_k$alternate between$f_1$and$f_2$, and the or we are in a "holding pattern" where$p_i$belong p_k$ belongs to $P$. A$or$B$and the two arrows out of$p_k$are both identities. Now it is not hard to convince yourself that given$(x, y)$in$A^2$or$B^2$, and$(x', y')$in$A^2$or$B^2$, if there is a zig-zag path from$x$to$x'$, and a zig-zag path from$y$to$y'$, then there is a zig-zag path from$(x, y)$to$(x', y')$with respect to the pair of maps $$A \times A \stackrel{f_1 \times f_1}{\leftarrow} P \times P \stackrel{f_2 \times f_2}{\to} B \times B.$$ B;$$all we do is pair together zig-zag paths in the separate x- and y-components. (Note that if the zig-zag to get from y to y' is longer than the zig-zag from x to x', we can always insert a holding pattern in the x-component so that the zig-zag in the y-component can "catch up", i.e., so that the lengths of the zig-zags match up.) But this means precisely that the map \phi is monic. 1 General pushouts, no; pushouts of pairs of epis, yes. In fact, yes to pushouts of pairs where only one of the arrows of the pair is epi. It's very easy to see the answer is no for general pushouts: since a coproduct A + B is the pushout of a pair A \leftarrow 0 \to B, and since the squaring functor preserves 0, the squaring functor would preserve this pushout only if it preserved the coproduct. But we all know that for finite cardinalities, a^2 + b^2 is generally not equal to (a+b)^2; therefore squaring cannot preserve coproducts. But in general, we can say that for the category of sets, given a pair of functions$$A \stackrel{f_1}{\leftarrow} P \stackrel{f_2}{\to} B,$$the canonical arrow$$\phi: A^2 +_{P^2} B^2 \to (A +_P B)^2$$is monic. (Lemma to be proved.) Assuming this, suppose for example that f_2: P \to B is epic. Then the pushout of f_2 along f_1 is also epic (well-known fact); let A \to C denote this epi (so C is shorthand for A +_P B). Then A^2 \to C^2 is also epic, and since this obviously factors as$$A^2 \to A^2 +_{P^2} B^2 \stackrel{\phi}{\to} C^2$$we see \phi is also epi. But it is monic by the lemma, hence \phi is an isomorphism, as we wanted to show. To prove the lemma, it helps to have a clear picture of how pushouts are formed in Set. The pushout C is the set of equivalence classes on A + B where x \in A + B is deemed equivalent to x' \in A + B iff there is a zig-zag path$$x = x_0 \stackrel{f_{i_1}}{\leftarrow} p_0 \stackrel{f_{i_2}}{\to} x_1 \leftarrow \ldots x_{n-1} \stackrel{f_{i_{n-1}}}{\leftarrow} p_{n-1} \stackrel{f_{i_n}}{\to} x_n = x'$$where the f_{i_k} alternate between f_1 and f_2, and the p_i belong to P. Now it is not hard to convince yourself that given (x, y) in A^2 or B^2, and (x', y') in A^2 or B^2, if there is a zig-zag path from x to x', and a zig-zag path from y to y', then there is a zig-zag path from (x, y) to (x', y') with respect to the pair of maps$$A \times A \stackrel{f_1 \times f_1}{\leftarrow} P \times P \stackrel{f_2 \times f_2}{\to} B \times B.$$But this means precisely that the map$\phi\$ is monic.